Question: A particle moves along the curve $x^2y^2=16$ so that the $x$ -coordinate is changing at a constant rate of $-2$ units per minute. What is the rate of change (in units per minute) of the particle's $y$ -coordinate when the particle is at the point $(1,4)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $-\dfrac{1}{4}$ (Choice C) C $\dfrac38$ (Choice D) D $8$
Explanation: Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This, however, shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dx}{dt}=-2$ for any value of $t$. We are asked for the rate of change of the particle's $y$ -coordinate when the particle is at the point $(1,4)$. In other words, we are asked for the value of $\dfrac{dy}{dt}$ at the point $(1,4)$. Finding $\dfrac{dy}{dt}$ $\dfrac{dy}{dt}=\dfrac{2y}{x}$ Finding $\dfrac{dy}{dt}$ at $(1,4)$ The expression for $\dfrac{dy}{dt}$ depends on both the particle's $x$ -coordinate ${1}$ and its $y$ -coordinate ${4}$ : $\begin{aligned} \dfrac{dy}{dt}&=\dfrac{2({4})}{({1})} \\\\ &=8\end{aligned}$ In conclusion, the rate of change of the particle's $y$ -coordinate when the particle is at the point $(1,4)$ is $8$ units per minute.